Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $k = \dfrac{-4q^2 - 36q}{q^2 - q - 90} \div \dfrac{q + 5}{q - 10} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{-4q^2 - 36q}{q^2 - q - 90} \times \dfrac{q - 10}{q + 5} $ First factor the quadratic. $k = \dfrac{-4q^2 - 36q}{(q - 10)(q + 9)} \times \dfrac{q - 10}{q + 5} $ Then factor out any other terms. $k = \dfrac{-4q(q + 9)}{(q - 10)(q + 9)} \times \dfrac{q - 10}{q + 5} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ -4q(q + 9) \times (q - 10) } { (q - 10)(q + 9) \times (q + 5) } $ $k = \dfrac{ -4q(q + 9)(q - 10)}{ (q - 10)(q + 9)(q + 5)} $ Notice that $(q + 9)$ and $(q - 10)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ -4q(q + 9)\cancel{(q - 10)}}{ \cancel{(q - 10)}(q + 9)(q + 5)} $ We are dividing by $q - 10$ , so $q - 10 \neq 0$ Therefore, $q \neq 10$ $k = \dfrac{ -4q\cancel{(q + 9)}\cancel{(q - 10)}}{ \cancel{(q - 10)}\cancel{(q + 9)}(q + 5)} $ We are dividing by $q + 9$ , so $q + 9 \neq 0$ Therefore, $q \neq -9$ $k = \dfrac{-4q}{q + 5} ; \space q \neq 10 ; \space q \neq -9 $